JEE Main 2024 (Online) 6th April Morning Shift | Statistics Question 4 | Mathematics

The mean and standard deviation of 20 observations are found to be 10 and 2 , respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

1.94

$$\sqrt{3.96}$$

$$\sqrt{3.86}$$

1.8

JEE Main 2024 (Online) 4th April Evening Shift

MCQ (Single Correct Answer)

-1

If the mean of the following probability distribution of a radam variable $$\mathrm{X}$$ :

$$\mathrm{X}$$	0	2	4	6	8
$$\mathrm{P(X)}$$	$$a$$	$$2a$$	$$a+b$$	$$2b$$	$$3b$$

is $$\frac{46}{9}$$, then the variance of the distribution is

$$\frac{581}{81}$$

$$\frac{566}{81}$$

$$\frac{151}{27}$$

$$\frac{173}{27}$$

JEE Main 2024 (Online) 4th April Morning Shift

MCQ (Single Correct Answer)

-1

Let $$\alpha, \beta \in \mathbf{R}$$. Let the mean and the variance of 6 observations $$-3,4,7,-6, \alpha, \beta$$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

$$\frac{16}{3}$$

$$\frac{11}{3}$$

$$\frac{14}{3}$$

$$\frac{13}{3}$$

JEE Main 2024 (Online) 1st February Evening Shift

MCQ (Single Correct Answer)

-1

Consider 10 observations $x_1, x_2, \ldots, x_{10}$ such that $\sum\limits_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum\limits_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. Then $\frac{\beta}{\alpha}$ is equal to :

$\frac{5}{2}$

$\frac{3}{2}$

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